History and Applications of the Natural Logarithm

Date: 03/02/98 at 11:45:35
From: Paula Fornwalt
Subject: history of e

We have been using e in continuous growth problems, where we are 
examining the growth potential of natural disease populations. I'm so 
surprised at how often this number comes up in other applications, 
though. Where did e come from? Who first derived it? Why is it so 
common in the field of biology?

Date: 03/02/98 at 12:32:09
From: Doctor Rob
Subject: Re: history of e

For its history, see the following URL:

Earliest Uses of Symbols for Constants, by Jeff Miller
  http://members.aol.com/jeff570/constants.html   

John Napier, the inventor of logarithms, is credited with discovering 
this constant. Leonhard Euler is credited with popularizing the use of 
the letter e for this number.

There are two reasons for its frequent appearance in mathematical 
contexts.

The first is that it is the limit, as n grows without bound, of

   (1 + 1/n)^n.

This particular definition lends itself to problems involving 
continuous compounding of interest. Think of adding 1/n of the current 
total to that current total n times.

The second is that e is the only constant such that the slope of the 
graph of y = e^x at 0 is 1; or, in other words, the function e^x is 
the solution of the differential equation dy/dx = y, with initial 
conditions y = 1 when x = 0. Similarly, the solution of dy/dx = a*y, 
with initial conditions y = 1 when x = 0, is e^(a*x).

This is the biological connection. You are talking about the 
instantaneous rate of growth of something being proportional to the 
amount of the something. That something can be a population of 
bacteria in a culture, or a population of cockroaches in a garbage 
dump. Here, "a" represents the fraction of them reproducing at any 
given time x. If "a" is negative, it could represent the fraction of 
organisms dying at any given time.

This is the model for radioactive decay of radium, uranium, and so on.
This is also the approximate model for inflation -- exponential 
growth.

The way to calculate e is by using as many terms as necessary for the
accuracy you need in the following infinite series:

   e = 1 + 1 + 1/2 + 1/(2*3) + 1/(2*3*4) + 1/(2*3*4*5) + 1/(2*3*4*5*6) 
         + ...

The terms shown above are already enough to obtain a value for e that 
is roughly 2.718.

-Doctor Rob, The Math Forum
 http://mathforum.org/dr.math/